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I am looking for three mappings f:X to Y any set of topology on X or Y. so very flexible. Can you help me find an example of a function that is (a) continuous but not an open or closed mapping (b) open but not closed or continuous (c) closed but not open of continuous

Thank you all

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HINT: If $X$ has the discrete topology, every function from $X$ to $Y$ is continuous, and every subset of $X$ is both open and closed. If you choose $Y$ so that it has a subset that isn’t open and a subset that isn’t closed, it’s not hard to get your first example. (You can even take $Y$ to be $X$ with a different topology.)

Suppose that $f:X\to Y$ is an open bijection; then it’s not hard to show that $f$ is also closed. Essentially the same argument shows that a closed bijection is always open. Thus, for the other two examples we cannot use bijections. It turns out that we can use an injection, though.

Let $X=\{0,1\}$ with the indiscrete topology, let $Y=\{0,1,2\}$, with the increasing nest topology

$$\tau=\big\{\varnothing,\{0\},\{0,1\},Y\big\}\;,$$

and let $f(0)=0$ and $f(1)=1$; this will work for one of the other two examples, and I’ll leave it to you to work out which one it is. Finally, you can use $X$ and $Y$ and a different injection from $X$ to $Y$ to get the remaining example.

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  • $\begingroup$ Thank you so much. As f(X)={0,1} in that example this maps an open set to an open set and there are no other open sets it must be an open mapping? and is not a closed mapping using the same logic as X is closed and it maps to an open set. sorry for the comment I am just trying to make sure my reasoning is okay. $\endgroup$ – Jm222222 Dec 10 '16 at 21:38
  • $\begingroup$ @Jm222222: You’re right about the open sets: the only open sets in $X$ are $\varnothing$ and $X$, and they both get sent to open sets in $Y$. However, you need to do a bit more to show that $f$ is not a closed map. It’s not enough to show that it takes a closed set to an open set, because that open set might also be closed. (Remember, a set can be both open and closed.) However, it’s true that $f[X]=\{0,1\}$ is not closed in this case, because $Y\setminus\{0,1\}=\{2\}$ is not open in $Y$. $\endgroup$ – Brian M. Scott Dec 10 '16 at 21:44
  • $\begingroup$ what is wrong with the function that maps the whole real line to a subset of R whose range is finite of numbers in the real line? If the target subset is a single value then thats its a continuous function, otherwise its not...right? $\endgroup$ – Pinocchio Jul 9 '18 at 19:02
  • $\begingroup$ Why is "If X has the discrete topology, every function from X to Y is continuous" correct? To my understanding in the discrete topology if $p$ is the limit point then the distance between $f(x)$ and $f(p)$ will always be 1 unless $f(x)=f(p)$. Therefore, every epsilon less than 1 won't work and thus the every function on the discrete metric is not continuous. Where did I go wrong? $\endgroup$ – Pinocchio Jul 10 '18 at 15:41

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